Question

pls give me the step by step explanation for this question ​

Answer 1

Answer:

-1

Step-by-step explanation:

2×{(sin^3x)^2 +(cos^3x)^2}-3{(sin^2x)^2 + (cos^2x)^2}

[•sin^2x+cos^2x=1]

=2×1-3×1

=-1

Answer 2

Answer:

4sinx cosx -1

Step-by-step explanation:

--->2(sin^6x + cos^6x)-3(sin^4x+cos^4x)

[a^3+b^3=(a+b)(a^2+b^2+ab)]

[a^2+b^2=(a+b)^2-2ab]

--->2(sin^2 x+cos^2 x)(sin^4x + cos^4x +sin^2x ×cos^2x) - 3[(sin^2 x+ cos^2 x )^2-2sinx cosx]

---->2×1(1-sinx cosx )-3(1-2sinx cosx)

---->2-2sinx cosx -3+6sinx cosx

---->4sinx cosx -1